Prove that the eigenvalues of an upper triangular matrix T are the entries of its main diagonal, so that the trace of T is the sum of its eigenvalues.
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Expert's answer
2013-02-25T12:17:15-0500
Eigenvalues are roots ofpolynomial det(T-kI)=0, where I is an identity matrix. Since T is upper triangular matrix, then we deal with determinant of upper triangular matrix again, and its determinant is product of diagonal entries. So, k is one of the roots of (a_11-k)(a_22-k)...(a_nn-k) = 0. Then all k varies over set {a_11,...,a_nn}, and sum of all k's is a_11+....+a_nn=tr(T).
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